Problem 17 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 20 , evaluate the line integral over the given curve. \(\int_{c} z d x+x d y+y d z ; C:\) the circular helix \(\mathbf{R}(t)=a \cos t^{\circ}+a \sin t^{\circ}+t k ; 0 \leq t \leq 2 \pi\)

Step-by-Step Solution

Answer

The value of the line integral is \(a^2 \pi\).

Step 1: Parameterize the curve

Given the curve \(\textbf{R}(t) = a \cos t \mathbf{i} + a \sin t \mathbf{j} + t \mathbf{k}\) for \(0 \leq t \leq 2 \pi \), parameterize the curve. Here, \(x = a \cos t\), \(y = a \sin t\), and \(z = t\).

Step 2: Find the derivatives

Compute the derivatives of the parametric equations: \(\frac{d x}{d t} = -a \sin t\), \(\frac{d y}{d t} = a \cos t\), and \(\frac{d z}{d t} = 1\).

Step 3: Express the differential terms

Express the differential forms \(dx\), \(dy\), and \(dz\) in terms of \(dt\). From Step 2, \(dx = -a \sin t \ dt\), \(dy = a \cos t \ dt\), and \(dz = dt\).

Step 4: Substitute and set up the integral

Substitute the expressions for \(x\), \(y\), and \(z\) as well as their differentials into the integral: \[\begin{aligned} \int_{c} z \ d x + x \ d y + y \ d z &= \int_{0}^{2 \pi} t \ (-a \ sin t) \ dt + \int_{0}^{2 \pi} (a \ cos t) \ (a \ cos t) \ dt + \int_{0}^{2 \pi} (a \ sin t) \ dt \end{aligned}\].

Step 5: Simplify and evaluate the integrals

Evaluate each integral separately: \[\begin{aligned} \int_{0}^{2 \pi} t (-a \ sin t) \ dt &= -a \int_{0}^{2 \pi} t \ sin t \ dt = 0, \int_{0}^{2 \pi} a^2 \ cos^2 t \ dt &= a^2 \int_{0}^{2 \pi} \left( \frac{1 + \cos(2t)}{2} \right) \ dt = a^2 \pi, \int_{0}^{2 \pi} a \ sin t \ dt &= a \left. -\cos t \right|_{0}^{2\pi} = 0. \end{aligned}\].

Step 6: Sum the results

Add the results of the integrals: \(-a \int_{0}^{2 \pi} t \sin t \ dt + a^2 \pi + 0 = a^2 \pi\). Thus, the value of the line integral is \(a^2 \pi\).

Key Concepts

parameterization of curves

When solving line integrals in vector calculus, one of the initial steps involves parameterization of the curve. This process allows you to express the curve in terms of a single parameter, typically denoted as \( t \). For example, in our exercise, the curve is a circular helix given by \( \mathbf{R}(t) = a \cos t \mathbf{i} + a \sin t \mathbf{j} + t \mathbf{k} \), where \( a \) is a constant, and \( 0 \leq t \leq 2\pi \).

Breaking this down:

• \( x = a \cos t \)
• \( y = a \sin t \)
• \( z = t \)

This representation translates the three-dimensional curve into a parameterized form based on \( t \), making it easier to compute integrals. By parameterizing, you convert complex spatial components into more manageable one-dimensional calculus.

vector calculus

Vector calculus is a key part of evaluating line integrals. It deals with fields in three dimensions by using vector functions. In our exercise, we found derivatives for the parameter functions to form differential expressions necessary for integration. The relations were:

• \( \frac{dx}{dt} = -a \sin t \)
• \( \frac{dy}{dt} = a \cos t \)
• \( \frac{dz}{dt} = 1 \)

Using these derivatives, we expressed the differential forms \(dx\), \(dy\), and \(dz\) in terms of the parameter \(dt\):

• \(dx = -a \sin t \ dt \)
• \(dy = a \cos t\ dt \)
• \(dz = dt \)

These differential forms simplify the integration process from vector fields along the curve.

evaluating integrals

The final step in solving the problem involves evaluating integrals. After substituting values into the integral \( \int_{c} z \, dx + x \, dy + y \, dz \), we ended up with three separate integrals:

• \( \int_{0}^{2\pi} t (-a \sin t) \, dt \)
• \( \int_{0}^{2\pi} (a \cos t)(a \cos t) \, dt \)
• \( \int_{0}^{2\pi} (a \sin t) \, dt \)

Each integral was evaluated separately:

• \( \int_{0}^{2\pi} t (-a \sin t) \, dt = 0 \)
• \( \int_{0}^{2\pi} a^2 (\cos^2 t) \ dt = a^2 \pi \)
• \( \int_{0}^{2\pi} a \sin t \, dt = 0 \)

Adding these results, the total value of the line integral was found to be \( a^2 \pi \). Proper parameterization and understanding vector functions are essential to breaking down and evaluating complex integrals effectively.